Problem: The lifespans of sloths in a particular zoo are normally distributed. The average sloth lives $15.7$ years; the standard deviation is $1.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a sloth living less than $18.1$ years.
Solution: $15.7$ $14.5$ $16.9$ $13.3$ $18.1$ $12.1$ $19.3$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $15.7$ years. We know the standard deviation is $1.2$ years, so one standard deviation below the mean is $14.5$ years and one standard deviation above the mean is $16.9$ years. Two standard deviations below the mean is $13.3$ years and two standard deviations above the mean is $18.1$ years. Three standard deviations below the mean is $12.1$ years and three standard deviations above the mean is $19.3$ years. We are interested in the probability of a sloth living less than $18.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the sloths will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the sloths will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $13.3$ years and the other half $({2.5\%})$ will live longer than $18.1$ years. The probability of a particular sloth living less than $18.1$ years is ${95\%} + {2.5\%}$, or $97.5\%$.